Question

Can you please help me, here is the problem?
The revenue R denied from selling x units of a commodity is computed as the product of x and the price P per unit.
a. If the price demand equation is p=1600-10x, where P is the price (in pesos) per units when x units are sold, find the revenue when 60 units are sold.
b. what is the maximum revenue in (a)?
c. if C is the cost, in pesos, of producing x units of the commodity then, assuming that every unit produced is sold, the profit P is given by P=R-C, revenue minus cost, if the cost of producing x units of commodity in (a) is C=975+60x, what is

Answer 1

help me please :( ?

Answer 2

please help me

Answer 3

thanks :)

Answer 4

a. substitute x=60 into the eqn p=1600-10x

Answer 5

\[p=1600-10(60)\]

Answer 6

so it will become p= 1600-600

Answer 7

yep,then subtract 1600 with 600.

Answer 8

the value of P=1000

Answer 9

yeap,that's right!

Answer 10

how about the problem b and c Sir?

Answer 11

what is the maximum revenue in a?

Answer 12

Can you please help me, here is the problem?
The revenue R denied from selling x units of a commodity is computed as the product of x and the price P per unit.
a. If the price demand equation is p=1600-10x, where P is the price (in pesos) per units when x units are sold, find the revenue when 60 units are sold.
b. what is the maximum revenue in (a)?
c. if C is the cost, in pesos, of producing x units of the commodity then, assuming that every unit produced is sold, the profit P is given by P=R-C, revenue minus cost, if the cost of producing x units of commodity in (a) is C=975+60x, what is the profit when 60 units are produced and sold?

Answer 13

Marc help me?

Answer 14

wait a minute,i think for question a,the question ask to find the revenue not the price.

Answer 15

ok thank your time

Answer 16

np...
equation for revenue
\[R=p(x)\]

Answer 17

eqn for price
\[p=1600-10x\]

Answer 18

so what is the value of our Revenue?

Answer 19

substitute p=1600-10x into R=px

Answer 20

what is the value of our x?

Answer 21

\[R=(1600-10x)(x)\]

Answer 22

in question a,it mention that 60 units are sold.

Answer 23

so,x is 60

Answer 24

sub. x=60 into the equation to find the revenue\[R=1600x-10x^2\]

Answer 25

so our minimum Revenue is 60,000

Answer 26

yep,our revenue is 60000.

Answer 27

how about the maximum Revenue how to solve it?

Answer 28

i'm x sure about question b but i know how to find for question c.

Answer 29

question c,
sub. x=60 into this eqn C=975+60x

Answer 30

\[C=975+60(60)\]

Answer 31

\[C=4575\]

Answer 32

show the solution marc, I have no idea how to solve it?

Answer 33

yea,that's what i'm doing right now... :)

Answer 34

okay for question C,they ask to find the profit when 60 units are produced and sold

Answer 35

given that the equation to find cost is\[C=975+60x\]

Answer 36

yes Marc,
then the formula for getting the profit is P=R-C?

Answer 37

yep,that's right

Answer 38

so the profit is P=55,425
because our Revenue is 60,000-Cost 4575 then the total is 55,425

Answer 39

yes,that's correct! ^

Answer 40

now lets proceed to problem b Marc?

Answer 41

here is the example Marc i think this one help you to solve problem b

Answer 42

Can you hep me with my question?

Answer 43

it seems that question b needs to use differentiation method

Answer 44

(1600-x)(975+60x) what do you think Marck?

Answer 45

do you think my representation is right, to get the Maximum revenue?

Answer 46

i think it should be\[R=(1600-10x)(x)\]

Answer 47

it is based on question a)
so,u need to use eqn for revenue and eqn for profit.

Answer 48

price*

Answer 49

i think so

Answer 50

show your proposal solution lets find out?

Answer 51

okay,here we go...\[R=1600x-10x^2\]

Answer 52

differentiate R

Answer 53

\[\frac{dR}{dx}=1600x^{1-1}-2(10)^{2-1}\]

Answer 54

i think (Price)(cost) is the formula to get the maximum value, what do you think so?

Answer 55

we do not need to involve cost since the eqn for revenue is R=px

Answer 56

a while ago we get the minimum value because the x is given at this case the x is unknown is in it?

Answer 57

for question a),it didn't say to get minimum value but revenue only.
for question b),they ask to find the maximum revenue means that dR/dx=0

Answer 58

\[\frac{dR}{dx}=1600-20x\]

Answer 59

when maximum revenue,dR/dx=0

Answer 60

where did you get your 20x?

Answer 61

how is it, where did you get that solution?

Answer 62

i differentiate it\[\frac{dR}{dx}=1600x^{1-1}-2(10)x^{2-1}\]

Answer 63

so what is the maximum value?

Answer 64

first we need to find the value of x(units for selling)\[\frac{dR}{dx}=1600-20x\]

Answer 65

when maximum revenue,dR/dx=0

Answer 66

\[0=1600-20x\]

Answer 67

find the value of x

Answer 68

20x=1600

Answer 69

x=80

Answer 70

the maximum value is 80

Answer 71

not yet,that is just for the value of x( units sold)

Answer 72

now,do the samething as question a.
Find the revenue using the value of x which is 80

Answer 73

show the final answer, because i need to take a rest it already midnight in our country

Answer 74

yea,me too... xD
Sorry,for taking ur time.
R=px
R=(price eqn)(x)
R=(1600-10x)(x)
R=(1600x-10x^2)
R=1600(80)-10(80)^2

Answer 75

Maximum revenue,R=?
solve it!

Answer 76

thanks Marc :)

Answer 77

i need to sleep now maybe we can continue our discussion tomorrow morning :D

Answer 78

thanks for your Greatness i learn alot :)

Answer 79

here bro

Answer 80

dRdx=1600x1−1−2(10)x2−1 how did you get this formula bro?